accuracy rate for that model? rates for k-fold cross-validation with k = 4, which model will we choose t Observations in the Validation Set 301-400 201-300 101-200 1-100 Model 1 69.5% 70.2% 71.8% 67.3% Model 2 72.2% 68.9% 71.5% 69.4%
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- In a multivariate model (ROC analysis), the area under the curve is 0.705 , explain how this is statistically significant. Images of the predictive power graph is attached. Please help explain the results.Consider all 40 observations on the delivery time data. Delete 10% (4) of the observations at random. Fit a model to the remaining 36 observations, predict the four deleted values, and calculate R^2 for prediction. Repeat these calculations 100 times. Calculate the-average R^2 for prediction. What information does this convey about the predictive capability of the model? How does the average of the 100R^2 for prediction values compare to R^2 for prediction based on PRESS for all 40 observations?What is the sensitivity of the following model? Actual Predicted yes no yes 40 8 no 2 50 Group of answer choices 0.8862 0.8621 0.9091 0.9523
- Why is RMSE generally the preferred performance measure for regression tasks? A. Because it gives an idea of the average percentage deviation of the predictions from the actual values. B. Because it gives an idea of the average absolute deviation of the predictions from the actual values. C. Because it gives an idea of how much error the system typically makes in its predictions, with a higher weight given to large errors. D. Because it gives an idea of the average deviation of the predictions from the actual values.Parametrize the monkey saddle z=x^3−3xy^2 by using u=x,v=y as the parameters in MATLAB. Then plot.Consider the following table that contains the IDs of 12 participants and their crash- avoidance reaction times (in milliseconds) on four tests. Two tests were done during the day in daylight and two tests were done at night with reduced lighting. Crash-avoidance Reaction Time (ms) Night Test 3 Participant Day ID Test 2 Test 1 Test 4 9001 887 838 765 648 9002 680 491 953 767 9003 662 553 985 696 9004 789 526 581 770 9005 508 451 688 714 9006 566 633 856 642 9007 656 747 846 718 9008 776 491 944 604 9009 770 672 814 617 9010 333 432 591 602 9011 730 593 796 700 9012 496 404 846 892 This question has multiple parts. Create output to MATLAB's command window exactly as shown, except replacing xxxx with actual values, as instructed. You may use the disp command, but not the fprintf command If you have difficulty with the alignment or formatting of numeric output, add the command format shortg near the top of your solution. In your solution, put the following MATLAB statements at the…
- You decide to run a simpler model to predict churn, using only the variables tenure (in months) and TotalCharges (in US$). The output is given below. The AIC of this model is 4727.6 (in contrast to the AIC of 4240 for the full model). On the basis of this which model would be expected to give superior predictive performance? Actual ## Coefficients: ## Estimate Std. Error z value Pr(>|z|) ## (Intercept) 2.471e-01 5.360e-02 4.611 4.01e-06 *** ## tenure < 2e-16 *** -1.124e-01 5.816e-03 -19.334 ## TotalCharges 8.236e-04 5.618e-05 14.660 < 2e-16 *** ## No --- ## Signif. codes: 0 ## Yes Yes ## Null deviance: 5701.5 on 4921 ## Residual deviance: 4721.6 on 4919 ## AIC: 4727.6 515 345 ## (Dispersion parameter for binomial family taken to be 1) ## Predicted ***** No 795 3267 0.001 Confusion Matrix (Training) **** Actual 0.01 Yes No degrees of freedom degrees of freedom Yes The simpler model (with just tenure and TotalCharges) The full model (with all variables) 0.05 0.1 220 145 Predicted No 339…In Reid and Jamieson (2023) the authors had to use a low learning rate in their Minerva 2 model to successfully account for DRM effects. Describe what the learning rate in Minerva 2 does . Using the results from the above simulation describe why you believe it is necessary to use a low learning rate when accounting for DRM results using distributional models .Arrange the steps executed in the k-nearest neighbor classification algorithm.
- The non-parametric density-based approach assumes that the density around a normal data observation within a cluster (relatively big) is similar to the density around its neighbours, and the density around an outlier (relatively small) is considerably different to the density around its neighbours. If we had the density of observation within a cluster smaller than the density around an outlier, explain why we would have such a situation. And provide a solution to this problem.How computationally hard is a full jackknife assessment of accuracy and variance for an unpruned nearest-neighbor classifier (Chap.??)?Given the following premises: (1) ∀x (R(x) → S(x))(2) ¬S(c) Use modus tollens to derive the conclusion ¬R(c). Show your step-by-step reasoning.